You titrated a 25.00 mL solution of 0.02 M oxalic acid with a freshly prepared solution...
You titrated a 25.00 mL solution of 0.02 M oxalic acid with a freshly prepared solution of KMnO4. If it took 41.81 mL of this solution to reach the endpoint, what was the molarity of the KMnO4?
0.0119 is wrong
+/- 3 sig figs.
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You titrated a 25.00 mL solution of 0.02 M oxalic acid with a
freshly prepared solution...
You titrated a 22.00 mL solution of 0.0300 M oxalic acid with freshly prepared solution of...
You titrated a 22.00 mL solution of 0.0300 M oxalic acid with
freshly prepared solution of KMnO4. If it took 48.99 mL of this
solution, what is the molarity of the KMnO4?
0.022 L of 0.300 M oxalic acid
(0.022)(0.03) = 0.00066 mol oxalic acid
0.0006 mol oxalic acid x (2 mol KMnO4)/(5 mol oxalic acid)=
0.000264 mol KMnO4
0.000264 mol/0.04899 L = 0.05388855 M KMnO4
= 0.005388 M KMnO4
where did I go wrong?
Question 7 0.1 / 1...
A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml. 25.00 ml of the...
A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml. 25.00 ml of the solution is taken out and is titrated to a neutral endpoint with 0.10 M NaOH. The titrated portion is then mixed with the remaining untitrated portion and the pH of the mixture is measured. Mass of acid weighed out (grams) 0.773 Volume of NaOH required to reach endpoint: (ml) 19.0 pH of the mixture Ihalf neutralized solution 3.54 Calculate the following...
A 25.00 mL solution containing 0.035 M sodium acetate is titrated with a 0.098 M solution...
A 25.00 mL solution containing 0.035 M sodium acetate is titrated with a 0.098 M solution of HCl. How many milliliters of HCI are required to reach the endpoint? Answer CHECK A 25.00 mL solution containing 0.035 M sodium acetate is titrated with a 0.098 M solution of HCI. How many milliliters of HCI are required to reach the endpoint? Answer: CHECK
A 0.100 M oxalic acid, HO2CCO2H, solution is titrated with 0.100 M KOH
A 0.100 M oxalic acid, HO2CCO2H, solution is titrated with 0.100 M KOH. Calculate the pHs when 25.00 mL of oxalic acid solution is titrated with 10, 15, 20, 25, 35, 40, 45, 50 and 55 mL of NaOH. Kal = 5.4 x 10-2 and Ka2 = 5.42 x 10-5 for oxalic acid. Show all your work and make a graph of pH versus Vb
A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml. 25.00...
A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml. 25.00 ml of the solution is taken out and is titrated to a neutral endpoint with 0.10 M NaOH. The titrated portion is then mixed with the remaining untitrated portion and the pH of the mixture is measured. Mass of acid weighed out (grams) 0.762 Volume of NaOH required to reach endpoint: (ml) 18.0 pH of the mixture (half neutralized solution) 3.15 Calculate the following...
16. Exactly 25.00 ml of 0.0685 M HClO4 solution is placed in a flask and titrated...
16. Exactly 25.00 ml of 0.0685 M HClO4 solution is placed in a flask and titrated with standardized 0.050M NaOH. A. Determine the endpoint of the titration B. Determine the pH of the solution after the following amounts of NaOH are added to the acid solution: a.) Initially, 0.00 ml NaOH. b.) 5.00 ml c.) 15.00 ml d.) 25.00 ml e.) 34.00 ml f.) At the endpoint of the titration g.)...
18. A 25.00 ml sample of the 50.00 ml solution was titrated with 0.1359 M potassium...
18. A 25.00 ml sample of the 50.00 ml solution was titrated with 0.1359 M potassium thiocyanate and 12.90 mL were required to reach the endpoint. Calculate the gram of silver (MM= 107.87 g/mol) in the sample.
50.00 mL of unknown calcium hydroxide solution is titrated with 0.300 M standard nitric acid solution....
50.00 mL of unknown calcium hydroxide solution is titrated with 0.300 M standard nitric acid solution. If 42.21 mL of the standard acid so lution is required to reach a phenolphthalein endpoint, what is the molarity of the unknown calcium hydroxide solution? (Hint: write the balanced formula unit equation.)
A 39.9 mL sample of 0.1721 M oxalic solution was titrated with 30.9 mL of sodium...
A 39.9 mL sample of 0.1721 M oxalic solution was titrated with 30.9 mL of sodium hydroxide solution according to the balanced equation. Calculate the molarity of the sodium hydroxide solution. 2NaOH + H2C2O4–>Na2C2O4+2H2O
Exactly 25.00 ml of a 0.0685 M pentanoic acid solution,(C4H8O2), is placed in a flask and...
Exactly 25.00 ml of a 0.0685 M pentanoic acid solution,(C4H8O2), is placed in a flask and titrated with standardized 0.050 M NaOH. (the pKa of this acid is 4.84). A. Determine the endpoint of the titration B. Determine the pH of the solution after the following amounts of NaOH are added to the acid solution: a.) Initially, 0.00 ml NaOH. b.) 5.00 ml c.) 15.00 ml d.) 25.00 ml e.) 34.00 ml...
You titrated a 22.00 mL solution of 0.0300 M oxalic acid with
freshly prepared solution of KMnO4. If it took 48.99 mL of this
solution, what is the molarity of the KMnO4?
0.022 L of 0.300 M oxalic acid
(0.022)(0.03) = 0.00066 mol oxalic acid
0.0006 mol oxalic acid x (2 mol KMnO4)/(5 mol oxalic acid)=
0.000264 mol KMnO4
0.000264 mol/0.04899 L = 0.05388855 M KMnO4
= 0.005388 M KMnO4
where did I go wrong?
Question 7 0.1 / 1...
A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml. 25.00 ml of the solution is taken out and is titrated to a neutral endpoint with 0.10 M NaOH. The titrated portion is then mixed with the remaining untitrated portion and the pH of the mixture is measured. Mass of acid weighed out (grams) 0.773 Volume of NaOH required to reach endpoint: (ml) 19.0 pH of the mixture Ihalf neutralized solution 3.54 Calculate the following...
A 25.00 mL solution containing 0.035 M sodium acetate is titrated with a 0.098 M solution of HCl. How many milliliters of HCI are required to reach the endpoint? Answer CHECK A 25.00 mL solution containing 0.035 M sodium acetate is titrated with a 0.098 M solution of HCI. How many milliliters of HCI are required to reach the endpoint? Answer: CHECK
18. A 25.00 ml sample of the 50.00 ml solution was titrated with 0.1359 M potassium thiocyanate and 12.90 mL were required to reach the endpoint. Calculate the gram of silver (MM= 107.87 g/mol) in the sample.
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